An electron travels in the +x direction with a velocity of 1.5 *10⁶ m/s. It enters an E * B velocity selector that utilizes a cross electric and magnetic field to produce a zero net force on the particle. The magnitude of the magnetic field is 0.10 T in the +y direction. The magnitude of the electric field required to produce a straight-line trajectory is A) 1.5 * 10⁵ V/m in the +z direction. B) 1.5 *0⁵ V/m in the -z direction. C) 1.5 * 10⁵ V/m in the +y direction. D) 1.5 * 10⁵ V/m in the -y direction.

Question

Quizplus · Accepted Answer

For a particle to move in a straight line in a velocity selector, the electric force must balance the magnetic force. The formula for the electric force is $F_e = qE$, and for the magnetic force, it's $F_m = qvB$, where $q$ is the charge of the particle, $v$ is its velocity, $E$ is the electric field strength, and $B$ is the magnetic field strength. Setting $F_e = F_m$, we get $E = vB$. Given $v = 1.5 \times 10^6 \, \text{m/s}$ and $B = 0.10 \, \text{T}$, $E = 1.5 \times 10^6 \times 0.10 = 1.5 \times 10^5 \, \text{V/m}$. The direction of the electric field is determined by the right-hand rule, given the velocity in the +x direction and the magnetic field in the +y direction, the electric field must be in the +z direction to balance the forces.

An Electron Travels in the +X Direction with a Velocity